Most Important Selected Qs for JEE AdvancedMathematicsDifferential Equations
If y satisfies the differential equation x d y=y (x y^3+1 ) d x & y(1)=2 , then -
Options
- A_ x x y^3=- 4 3
- B_ x y=0
- Ccurve y=f(x) is symmetric w.r.t. origin
- Dcurve y =f( x ) is continuous x R .
Correct answer
C. curve y=f(x) is symmetric w.r.t. origin
Step-by-step solution
x d y d x -y=x y^4 Dividing both sides by y ^4 & putting 1 y ^3 = t - x 3 dt dx - t = x or dt dx + 3 t x =-3 which on solving gives x^3 y^3 = -3 x^4 4 +c y(1)=2 c= 7 8 Hence y ^3= 8 x ^3 7-6 x ^4 f( x )= [3] 8 x ^3 7-6 x ^4 which is odd and discontinuous at point where 7-6 x^4=0